Download Chapter 2 (Complex Numbers)

2
Chapter
Complex Numbers
Chapter Outline
2.1
Introduction and Definitions
2.2
Argand Diagram
2.3
The Modulus and Argument of a Complex Numbers
2.4
The Polar Form of Complex Numbers
2.5
The Exponential Form of Complex Numbers
2.6
De Moivre’s Theorem
2.7
Finding Roots of Complex Numbers
2.8
Expansion for cos n and sin n in terms of cosines and sines of multiple of θ
2.9
Loci in the complex numbers
Tutorial 2
1
2.1 INTRODUCTION AND DEFINITIONS
A collection of real number system was not complete. We realize that a quadratic
equation a 2 x  bx  c  0 , could not be solved if b2 – 4ac <0. Consider a quadratic
equation x 2  1  0, a  c  1, b  0 then we have x    1 and x    1 , Since we
know that the square of any real number must be greater than or equal than zero, we
realize that the real number system was not complete. This leads to a study of complex
numbers which are useful in variety of applications. A complex number system allowed
the possibility of negatives squares and had the real numbers as a subset, R. In above
quadratic equation we do not have any problems in solving this equation as complex
number system allowed a negative square. We introduce a symbol, j , with the property
that j 2  1 and it follows j   1 . We can use this to write down the square root of
any negative numbers. Then the solution of the equation are given by x = j and x = -j.
9 and  9 .
Example: Write down the expression of the
Solution:
9 3
 9  9  1  9   1  3   1  3 j
Simplify j 3 , j 4 , j 8
Definition 2.1: Complex Number
If z is a complex number then we write
z  a  bj
where a, b Є R , a is a real part and b is the imaginary part.
2
Example 1
Express in the form z  a  bj .
a)
 25
c) 5 +
 36
b)
 20
d)
27 +
 18
Solution:
25  1 = 5 j
a)
 25 =
25(1) =
b)
 20 =
4(5)( 1) = 2 5 j
c) 5 +
d)
 36 = 5  6 j
27 +
 
 18 = 3 3  3 2 j
Definition 2.2: Equal Complex Numbers
Two complex numbers z1  a  bj and z2  c  dj are equal if and only if a = c and b = d.
Example 2
Given 2 x  3 yj  4  3 j. Find x and y.
Solution:
x = 2 and y = 1
Definition 2.3: Operations of complex numbers
If z1  a  bj
and
z 2  c  dj then,
i) z1  z2  (a  c)  (b  d ) j
ii) z1  z2  (a  c)  (b  d ) j
iii) z1z2  (a  bj)(c  dj )
 ac  adj  bcj  bdj 2
 (ac  bd )  (ad  bc) j
3
Example 3
Given z1  2  3 j and z 2  1  j . Find
a) z1  z2
b) z1  z2
c) z1 * z2
d) Determine the value of z  73  4 j   2  j 5  6 j 
Solution:
a) z1 + z2 = (2 + 3j) + (1 – j)= (2 + 1) + (3 – 1)j= 3 + 2j
b) z1 – z2 = (2 + 3j) - (1 – j)= (2 - 1) + (3 + 1)j= 1 + 4j
c) z1 * z2 = (2+3j)(1-j) = 2 – 2j + 3j -3j2 = 5 + j
d) z  7 3  4 j   2  j 5  6 j 

 21  28 j   10  5 j  1 j  6 j 2
 21  28 j   10  6  5  12 j 

 21  28 j   16   7 j 
 21  16  28  7  j
 5  35 j
Definition 2.4: Complex Conjugate
If z  a  bj, its complex conjugate, denoted by z , is
z  a  bj
* Show that z z  a 2  b2
Example 4
Find the complex conjugate of:
a) z = 4 + 5j
b) z = 4 - 5j
c) z   j
d) z = j
4
Solution:
a) z = 4 - 5j
b) z = 4 + 5j
c) z = j
d) z = -j
Properties of conjugate complex number:
a) z  z
b) z1  z 2  z1  z 2
c) z1  z 2  z1  z 2
d) z1  z 2  z1  z 2
z  z
e)  1   1
 z2  z2
f)
1 1
 
z
z
i)
zz
 Im( z )
2
h)
zz
 Re( z )
2

g) z n  z
Definition 2.5: Division of complex numbers
If z1  a  b
and
z 2  c  d then
z1 a  bj

z 2 c  dj
a  bj c  dj

c  dj c  dj
ac  bd  (bc  ad ) j

c2  d 2

Example 5:
Determine the value of:
a)
2 j
1 j
b)
1 j
3 j
c) z 
7
1  j2

3 j 3 4 j
5
n
: n integer
Solution:
a)
2 j 2

1  j  1 
1 j 1
b)

3  j  3 
c) z 


j  1 

j  1 
j  2  2 j  j  j2 1 3
  j

j 
1  j2
2 2
j  3 

j  3 
j  3  j  3 j  j2 1 2
  j

j 
9  j2
5 5
7
1  j2

3 j 3 4 j
1  j 23  4 j 
73  j 

3  j 3  j  3  4 j 3  4 j 
21  7 j 

3  j 6  4 j  8 j 
2
9  j3  3 j  j  9  12 j  j12  16 j 
2
2
21  7 j   3  8  j10
9  1
9  16
 2.1  j (0.7)   0.2  j (0.4) 

 2.3  1.1 j
2.2 ARGAND DIAGRAM
The complex number z  a  bj is plotted as a point with coordinates (a,b). Such
a diagram is called an Argand diagram.
Figure 1
6
Example 6
Plot the following complex numbers on an Argand diagram.
a) 3j
b) 5 + 2j
c) -4
d) -1 – 2j
Solution:
y
3 ● (0,3)
2
● (5,2)
1
(0,-4)
●
-4 -3
-2
-1
0
1
2
3
4
5
6
-1
(-1,-2) ●-2
Example 7
Given that z1  1  3 j and z2  3  j . Plot z1  z2 in an Argand diagram.
Solution:
y
4
(4,4)
3
(1, 3)
z1  z2
2
1
(3,1)
0
1
2
3
4
x
5
7
x
2.3 THE MODULUS AND ARGUMENT OF A COMPLEX NUMBERS
Consider an Argand diagram in Figure 2, the distance of the point (a,b) imaginary
axis and real axis.
Figure 2
Definition 2.6: Modulus of complex numbers
The modulus of a complex number z = a + bj written as z is
r= z =
a 2  b2
Example 8
Find the modulus of the following complex numbers.
a) 3 – 4j
b) 5 + 2j
c) -1 – j
d) 7j
Solution:
a) |3 – 4j| = 3 2  (4) 2 =
b) |5 + 2j| = 52  2 2 =
25 = 5
29
c) |-1 – j| = 2
d) |7j| = 7
8
Properties of Modulus
a)
| z || z |
b) zz | z |2
z1 | z1 |

, z2  0.
z2 | z2 |
c) | z1 z2 || z1 || z2 |
d)
e) | z n || z |n
f) | z1  z2 || z1 |  | z2 |
g) | z1  z2 || z1 |  | z2 |
Definition 2.7: Arguments of complex numbers
The argument of the complex number, z = a + bj often given the symbol θ is define as
b
a
  tan 1   .
Figure 3
Example 9
Find the arguments of the following complex numbers;
a) 3 – 4j
b) 5 + 2j
c) -1 – j
d) 1 + 7j
Solution:
If θ is the argument of the following complex numbers, therefore,
 4
a) θ = tan-1 
 = -53.13o = 306.87o
 3 
9
2
b) θ = tan-1   = 21.80o
5
  1
c) θ = tan-1   = 225o
  1
7
d) θ = tan-1   = 81.87o
1
2.4 THE POLAR FORM OF COMPLEX NUMBERS
Figure 4
Using trigonometry we can write,
sin  
b
; b  r sin 
r
cos  
a
; a  r cos 
r
tan  
b
b
;   tan 1  
a
a
and;
|z| = r =
a 2  b2
Therefore, we can then write z = a + bj in polar form as:
z = r cos θ + j r sin θ
= r (cos θ + j sin θ)
or
= (r, θ)
10
Example 10
State the complex numbers of z when
b) z = 3 – 4j
a) z = 1 + j
Solution:
a) r = 12  12 = 2 and θ = tan 1 1 = 45o
Hence, z =
b) r =
2 (cos 45o + j sin 45o) or z = ( 2 ,45o)
 4
3 2  (4) 2 = 5 and θ = tan 1 
 = - 53.13o @ 306.87o
 3 
Hence, z = 5(cos 306.87o + j sin 306.87o) or z = (5, 306.87o)
Exercise 11
State the following complex numbers into the form of z = a + bj.
a) z = 4(cos 30o + j sin 30o)
b) z = 3 (cos 135o + j sin 135o)
Solution:
a) r = 4 dan θ = 30o
 3
= 2 3
a = r cos θ = 4 cos 30o = 4

2


1
b = r sin θ = 4 sin 30o = 4  = 2
2
Hence, z = 4(cos 30o + j sin 30o) = 2 3  2 j
b) r =
3 dan θ = 135o
a = r cos θ =
3 cos 135o =
b = r sin θ =
3 sin 135o =
Hence, z =
3
 1 
3 
= 
2
2

 1 
3
=
 2
3
2
3 (cos 135o + j sin 135o) = 
11
3
3

j
2
2
Definition 2.8:
If z1 and z2 are two complex numbers in polar
z1  r1 (cos 1  j sin 1 ) and z2  r2 (cos  2  j sin 2 ) therefore
form
i) z1z 2 = r1r2 [cos(1   2 )  j sin( 1  2 )]
r
z
ii) 1 = 1 [cos(1   2 )  j sin( 1   2 )]
r2
z2
Example 12
Given that z1 = 4(cos 30o + j sin 30o) and z 2 = 2(cos 60o + j sin 60o). Find
z
a) z1z 2
b) 1
z2
Solution:
a) z1z 2 = 4.2 [cos (30o + 60o) + j sin (30o + 60o)]
= 8 (cos 90o + j sin 90o)
=8j
z
4
b) 1 = [cos (30o - 60o) + j sin (30o - 60o)]
z2 2
= 2 [cos (-30o) + j sin (-30o)]
 3 1 
 j 
= 2
2
2 

= 3 j
2.5 THE EXPONENTIAL FORM OF COMPLEX NUMBERS
Definition 2.9:
Complex number z = r(cos θ + j sin θ) can also be written in exponential or
Euler form as z = reθj where θ is measured in radians and eθj = cos θ + j sin θ.
12
where
Example 13
State the following angles θ in radians.
a) 45o
b) 120o
c) 270o
Solution:
a) 450  450.

180
b) 120 0  120 0.
0


180
c) 270 0  270 0.

4
0

2
3
0

3
2

180
Example 14
Express the complex number z = 1 + j in exponential form.
Solution:
From Example 10 (a), we know that z = 1 + j =
r = 2 dan θ = 45o.
Then we have 450 
Hence, z = 1 + j =
 2e

4
.
2 (cos 45o + j sin 45o)

j
4
Definition 2.10:
If z1  r1e1 j and z2  r2 e2 j , then
i) z1z 2 = r1r2 e (1 2 ) j
z
r e (1 2 ) j
ii) 1 = 1
, z2  0
r2
z2
13
2 (cos 45o + j sin 45o) which
Example 15


j
j
If z1  8e 2 and z2  2e 3 , find
a) z1 z 2
z1
z2
b)
Solution:
a) z1 z 2  8.2e
  
  j
2 3
  
  j
2 3
z1 8e

b)
z2
2
 16e

 4e 6
5
j
6
j
2.6 DE MOIVRE’S THEOREM
Definition 2.11:
If z = r(cos θ + j sin θ) is a complex number in polar form to any power n, then
zn = rn(cos nθ + j sin nθ)
with any value n.
Example 16
If z = cos 30o + j sin 30o, find z3 and z5.
Solution:
a) z3 = (cos 30o + j sin 30o)3
= cos 3(30o)+ j sin 3(30o)
= cos 90o + j sin 90o
=j
b) z5 = (cos 30o + j sin 30o)5
= cos 5(30o) + j sin 5(30o)
= cos 150o + j sin 150o
= 
3 1
 j
2
2
14
Example 17
If z = 2(cos 25o + j sin 25o). Calculate
2
a) z5
b) z-3
c) z 3
Solution:
a) z5 = [2(cos 25o + j sin 25o)]5
= 25[cos 5(25o) + j sin 5(25o)]
= 32(cos 125o + j sin 125o)
= 32(-0.5736 + 0.8192j)
= -18.355 + 26.214j
b) z-3 = 2-3(cos -75o + j sin -75o)
=
1
(0.2588 – 0.9659j)
8
= 0.0324 – 0.1207j
2
2
c) z 3 = 2 3 [cos
2
2
(25o) + j sin (25o)]
3
3
= 1.587( cos 16.67o + j sin 16.67o)
= 1.52 + 0.455j
2.7 FINDING ROOTS OF COMPLEX NUMBERS
Definition 2.12:
If z = r(cos θ + j sin θ) then, the n root of z is
1
n
1
z = r n (kos
  360o k
  360o k
 j sin
) if θ in degrees
n
n
or
1
n
1
z = r n ( kos
  2k
  2k
 j sin
) if θ in radians
n
n
for k = 0, 1, 2, ...., n – 1.
15
Example 18
Find the cube roots of z = 8j.
Solution:
z = 8j or in polar form z = 8(cos 90o + j sin 90o) then
90o  360o k
90o  360o k
 j sin
) for k = 0, 1, 2.
z = 8 (cos
3
3
1
3
1
3
when k = 0
1
z 3 = 2(cos 30o + j sin 30o)
=
3 j
when k = 1
1
3
z = 2(cos 150o + j sin 150o)
=  3 j
when k = 2
1
z 3 = 2(cos 270o + j sin 270o)
= -2j
Therefore the roots of z = 8j are
3  j ,  3  j and -2j.
2.8 EXPANSIONS FOR COSn AND SINn IN TERMS OF COSINES AND
SINES OF MULTIPLES OF θ
Definition 2.13:
If z = cos θ + j sin θ then
i) z n 
1
= 2cos nθ
zn
ii) z n 
1
= 2j sin nθ
zn
16
Definition 2.14: Binomial Theorem
If n  N , then
( a  b) n  a n  nC1a n 1b nC2 a n 2 b 2  ... nC r a n r b r  ... nCn 1ab n 1  nCn b n
n
with Cr 
n!
.
r!( n  r )!
Example 19
State sin5 θ in terms of sines
Solution:
Using theorem 4, z = cos θ + j sin θ , then
1
z  = 2j sin θ
z
1
(2j sin θ)5 = [ z  ]5 .
z
By using Binomial expansion, then
2
3
4
1
1
1
1 1
1
[ z  ]5 = z5 – 5z4   + 10z3   - 10z2   + 5z   -  
z
z
z
z z
z
1  
1 
1


=  z 5  5  - 5  z 3  3  +10  z  
z
z  
z 


Thus,
32 j sin5 θ = 2j sin 5θ – 5(2j sin 3θ) + 10(2j sin θ)
sin5 θ =
1
5
5
sin 5θ sin 3θ + sin θ
16
16
8
17
5
2.9 LOCI IN THE COMPLEX NUMBERS
Definition 2.15:
A locus in a complex plane is the set of points that have a specified property. A
locus of a point in a complex plane could be a straight line, cirle, ellipse and etc.
Example 20
If z = a + bj, find the equation of the locus defined by:
i)
z2j
1
z 1
ii) z  (2  3 j )  2
Solution:
i)
z2j
a  bj  2 j

1
z 1
a  bj  1
a  bj  2 j  a  bj  1
a  (b  2) j  a  1  bj
a 2  (b  2) 2  ( a  1) 2  b 2
a 2  b 2  4b  4  a 2  2a  1  b 2
2a  4b  3  0
The locus is a straight line with slope
1
.
2
ii) z  (2  3 j )  a  bj  2  3 j )  2
a  2  (b  3) j  2
( a  2) 2  (b  3) 2  2
(a  2) 2  (b  3) 2  2 2
The locus is a circle of radius 2 and centre (2,3).
18
Tutorial 2
1. Simplify the following (addition and subtraction):
a) (3  j )  (1  2 j )
c) (2  3 j )  (1  2 j )
b) (5  3 j )  (4  3 j )
d) (1  j )  (1  j )
2. Simplify the following (multiplication):
a) (2  3 j )(4  5 j )
e) (u  vj)(u  vj) ,
b) (2  j )(3  2 j )
f) (x  2 yj )(2x  yj )
c) (1  j )(1  j )
g) j( 2 p  3 pj)
d) (3  4 j )(3  4 j )
h) (p  2qj )(p - 2qj )
3. Simplify the following (division):
a)
1- j
1 j
e)
1
x  yj
b)
1
2-3 j
f)
1
x-yj
c)
3 j-2
1 2 j
g)
1
1

23j 23j
d)
5 4 j
5-4 j
4. Simplify the following:
1
,
(1  j ) 3
a) (4  5 j ) 2
c)
b) (1  j ) 3
d) (cos   j sin  )(cos   j sin  )
19
5. Find the value of a and b.
a)
if (a  b)  j (a - b)  (2  5 j ) 2  j (2  3 j )
b)
Given that, (a  b)  j (a - b)  (1  j ) 2  j (2  j )
6. Evaluate the following
a) (2  3 j )(6  7 j )
b) (2  3 j )( 2  3 j )
c)
43j
2 j
7. Write in the simplest form:
a) (5  4 j )(3  7 j )( 2  3 j )
b)
(2  3 j )(3  2 j )
(4  3 j )
8. State the following in the form of a  bj .
a)
23j
2

j (4  5 j ) j
b)
1
1

2  3 j 1 2 j
9. If z 
2 j
1
, state z  in complex number.
z
1 j
10. Find the complex number which satisfies the equation:
3z z  2( z  z )  39  12 j
11. Label the following complex numbers on Argand Diagram.
a) 2  j
c) 3 j
b)  1  2 j
20
12. Find the modulus and argument of the following complex numbers:
a) 3  4 j
d) 5  12 j ,
b) 4 j
e)  1 2 j ,
c)  2  5 j
13. State the following in polar form:
a) 5  5 j
c) (4  7 j )( 2  3 j )
b)  6  3 j
d)
43j
2 j
14. State the following in exponential form:
a) 1  j
b) 1  3 j
c) 4(cos 30   j sin 30  )
Answers
1. a) (4  3 j )
b) 9
c) (1  5 j )
d) 2 j
2. a) - 7  22j
e) u 2  v 2 ,
b) 8  j
f) 2(x 2  y 2 )  5xyj ,
c) 2
g)  3 p  2 pj ,
d) 25
h) p 2  4q 2
21
3. a)  j
4.
e)
x
y
 2
j
2
x y
x  y2
2
b)
2
3

j
13 13
f)
x
y
 2
j
2
x y
x  y2
c)
4 7
 j
5 5
g)
4
13
d)
9 40

j
41 41
2
a) - 9 - 40j
b) - 2 - 2j
c)
1 j
4
d) 1
b) a 
a  2; b  20
5.
a)
6.
a)  9  32 j
b) 13
c) 1  2 j
7.
a) 115  133 j
b)
63 16

j
25 25
8.
a)
22 75

j
41 41
b)
23 11

j
65 65
9.
7
9

j
10 10
10. x  2; y  3
22
3
5
;b  
2
2
11. a) (2,1)
b) (1,2)
c) (0,3)
12. a) z  5,  tan 1  43 
b) z  4,   tan 1 ()  2
c) z  3,  tan 1
 
5
2
d) z  13,   tan 1  512 
e) z  5 ,   tan 1 2 ,
13. a) z  5 2 (cos 45 0  j sin 45 0 )
b) z  3 5 (cos 153.4 0  j sin 153.4 0 ) ,
c) z  31(cos 356 o  j sin 356 o )
d) z  5 (cos 63o  j sin 63o ) ,
5
14.
a) z  2e 4
5
b) z  2e 2
j
j
 j
c) z  4e 6
S. Asma, A. M. Fadzilah, M. Ibrahim. (2012). College Mathematics. Shah Alam,
Selangor: IPTA Publication Sdn. Bhd.
H. A. Abdull, K. J. Ahmad, S. Elyana, Z. Haslina, A. M. K. Khairul, S. Shakila, M. Z.
Shazalina, R. G. A. Siti, N. M. A. Wan. (2010). Teaching Module: Intermediate
Mathematics. Perlis: Penerbit UNIMAP.
C2. Yang, C. F. Wong, S. Jay. (2009). College Matriculation Mathematics QS026
Semester II. Shah Alam, Selangor: SAP Publications (M) Sdn. Bhd.
23